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JEE Advanced 2016Paper 1
Exam
JEE Advanced
Year
2016
Paper
Paper 1
Subject
Mathematics
Chapter
Binomial Theorem

JEE Advanced 2016 · Paper 1 · Mathematics · Binomial Theorem

Let m be the smallest positive integer such that the coefficient of...

Let m be the smallest positive integer such that the coefficient of x2{x}^{2} in the expansion of (1+x)2+(1+x){\left(1 + x\right)}^{2} + {\left(1 + x\right)}^{3} + \ldots + {\left(1 + x\right)}^{49} + {\left(1 + mx\right)}^{50}is(3n+1)51C3is \left(3n + 1\right){}^{51}{C}_{3} for some positive integer n.
Then the value of n is

Correct answer

0

Step-by-step solution

Coefficient of x2{x}^{2} in the expansion of
(1+x)2+(1+x)3+…(1+x)49+(1+mx)50{\left(1 + x\right)}^{2} + {\left(1 + x\right)}^{3} + \ldots {\left(1 + x\right)}^{49} + {\left(1 + mx\right)}^{50} is
2C2+3C2+…49C2+50C2m2=(3n+1)51C3{}^{2}{C}_{2} + {}^{3}{C}_{2} + \ldots {}^{49}{C}_{2} + {}^{50}{C}_{2}{m}^{2} = \left(3n + 1\right){}^{51}{C}_{3}
⇒ 3C3+3C2+…49C2+50C2m2=(3n+1)51C3{}^{3}{C}_{3} + {}^{3}{C}_{2} + \ldots {}^{49}{C}_{2} + {}^{50}{C}_{2}{m}^{2} = \left(3n + 1\right){}^{51}{C}_{3} (Use nCr+nCr+1=n+1Cr+1{}^{n} {C}_{r} {+}^{n} {C}_{r + 1} {=}^{n + 1} {C}_{r + 1})
⇒ 50C3+50C2m2=(3n+1)51C3{}^{50}{C}_{3} + {}^{50}{C}_{2}{m}^{2} = \left(3n + 1\right){}^{51}{C}_{3}
50.49.486+50.492m2=(3n+1)51.50.496\dfrac{50.49.48}{6} + \dfrac{50.49}{2}{m}^{2} = \left(3n + 1\right)\dfrac{51.50.49}{6}
m2=51n+1{m}^{2} = 51n + 1 must be a perfect square
By trial ⇒ n = 5andandm = 16((M, n \in N)<br/>)<br />⇒ n = 5
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